Question

My assignment says to do the following:
Search2: Search for a solution to x*x + y*y – 12x -10y + 36 = 0. Search from 0 to 10 in both x and y, searching every y value before moving to the next x. Print the first three solutions found. (Note – a labelled break is handy here!)

I can’t figure out the logic for this. I think I have to use more than 2 loops but not sure.
This is what I have so far (It just repeats (6,0)):

for (int j = 0; j <= 10; j++) {
    for (int i = 0; i <= 10; i++) {
        while (((i * i) + (j * j) - (12 * i) - (10 * j) + 36) == 0) {
            System.out.println("(" + i + ", " + j + ")");  
        }  
    }  
}  

UPDATE
Here is the solution:

    int t = 0;

    for (int i = 0; i <= 10; i++) {
        if (t == 3) {
            break;
        }
        for (int j = 0; j <= 10; j++) {
            if (((i * i) + (j * j) - (12 * i) - (10 * j) + 36) == 0) {
                System.out.println("(" + i + ", " + j + ")");
                t++;
            }
        }
    }

Answer 1

Not a bad attempt. Because you’re so close, I’ll show you a working solution. Basically, you need to do three things:

1. Change while to if
2. Use a variable to count the number of times you find the solution, so you can stop at three
3. Add a label so you can break out of the outer loop from within the inner loop

I also recommend you use variable names the same as the problem – ie x and y – for clarity.

int count = 0;
outerLoop:
for (int y = 0; y <= 10; y++) {
    for (int x = 0; x <= 10; x++) {
        if (x * x + y * y - 12 * x - 10 * y + 36 == 0) {
            System.out.println("(" + x + ", " + y + ")");  
            if (++count == 3)
                break outerLoop;
        }
    }
}

When executed, this code produces:

(6, 0)
(3, 1)
(9, 1)

Sorry for spoon-feeding you, but part of the lesson here is good coding style and practice.